An intergal identity involving two kinds of Bessel functions I am working on a problem involving Bessel functions and interested in the following identity 
$$
\int_0^\infty \frac{J_\nu(a x) Y_\nu(b x)-J_\nu(b x) Y_\nu(a x)}
{x(J_\nu(bx)^2+Y_\nu(bx)^2)}dx = 
-\frac{\pi}{2}(\frac{b}{a})^\nu
$$
where $J$ is the first kind Bessel function and $Y$ is the second kind Bessel function.
However, I found this identity in "Table of Integrals, Series, and Products", which has no explanation. 
And the reference therein provides no detail and reference neither.
I hope to find more detail on how to deriving such result. 
Any reference and suggestion would be greatly appreciated! 
 A: We suppose $a,b>0$. First, as the Hankel function (DLMF)
\begin{equation}
{H^{(1)}_{\nu}}\left(z\right)=J_{\nu}\left(z\right)+iY_{\nu}\left(z\right)
\end{equation} 
it can be recognized that 
\begin{equation}
\frac{J_\nu(a x) Y_\nu(b x)-J_\nu(b x) Y_\nu(a x)}
{J_\nu(bx)^2+Y_\nu(bx)^2}=-\Im \left[\frac{H_\nu^{(1)}(ax)}{H_\nu^{(1)}(bx)}\right]
\end{equation} 
Using the parity properties of the Hankel functions, it can be shown that
\begin{align}
\Re \left[\frac{H_\nu^{(1)}(axe^{i\pi})}{H_\nu^{(1)}(bxe^{i\pi})}\right]&=\Re \left[\frac{H_\nu^{(1)}(ax)}{H_\nu^{(1)}(bx)}\right]\\
\Im \left[\frac{H_\nu^{(1)}(axe^{i\pi})}{H_\nu^{(1)}(bxe^{i\pi})}\right]&=-\Im \left[\frac{H_\nu^{(1)}(ax)}{H_\nu^{(1)}(bx)}\right]
\end{align}
Then,
\begin{align}
I&=\lim_{\epsilon\to0^+}\int_\epsilon^\infty \frac{J_\nu(a x) Y_\nu(b x)-J_\nu(b x) Y_\nu(a x)}
{x(J_\nu(bx)^2+Y_\nu(bx)^2)}\,dx\\
&=-\frac{1}{2i}\lim_{\epsilon\to0^+}\left[\int_{-\infty}^{-\epsilon} \frac{H_\nu^{(1)}(ax)}{H_\nu^{(1)}(bx)}\,\frac{dx}{x}+\int_{\epsilon}^\infty \frac{H_\nu^{(1)}(ax)}{H_\nu^{(1)}(bx)}\,\frac{dx}{x}\right]
\end{align}
We close the contour a semi-circular path, centred at the origin and avoiding the origin from above to express
\begin{equation}
-2iI+I_R+I_\epsilon=0
\end{equation} 
as no poles exist in the upper half-plane. Here, $I_R$ corresponds to the contribution of the large semi-circular path and $I_\epsilon$ to that of the small one near the origin. We observe that for $x\to \infty$ in the upper half-plane, 
\begin{equation}
\frac{H_\nu^{(1)}(ax)}{H_\nu^{(1)}(bx)}\sim \sqrt{\frac{b}{a}}e^{i(a-b)x}\to 0
\end{equation} 
if $a>b$. Then the semi-circular path contribution to the integral vanishes. The small clock-wise half circular contribution near the origin is $$I_\epsilon=-i\pi \lim_{x\to 0}  \frac{H_\nu^{(1)}(ax)}{H_\nu^{(1)}(bx)}$$ It can be evaluated using the representations of $H^{(1)}_{\nu}\left(z\right)$ and of $J_\nu(z)$: 
\begin{align}
{H^{(1)}_{\nu}}\left(z\right)&=i\csc\left(\nu\pi\right)\left(e^{-\nu\pi i}J_{\nu}\left(z\right)-J_{-\nu}\left(z\right)\right)\\
J_{\nu}\left(z\right)&=(\tfrac{1}{2}z)^{\nu}\sum_{k=0}^{\infty}(-1)^{k}\frac{(
\tfrac{1}{4}z^{2})^{k}}{k!\Gamma\left(\nu+k+1\right)}
\end{align} 
We obtain
\begin{equation}
I_\epsilon=\begin{cases}-i\pi\left( \frac{b}{a} \right)^\nu \text{ for } \nu>0\\
-i\pi\left( \frac{a}{b} \right)^\nu \text{ for } \nu<0
\end{cases}
\end{equation} 
Finally
\begin{equation}
I=-\frac{\pi}{2}\left( \frac{b}{a} \right)^{\left|\nu\right|}
\end{equation} 
for $0<b<a$ as tabulated in G&R (6.542), but the condition on the sign of $\nu$ is not given (at least in the 5th edition).
